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Published by Perpus Kota Semarang, 2018-10-09 19:49:19

engineering-fluid-mechanics

engineering-fluid-mechanics

T. Al-Shemmeri

Engineering Fluid Mechanics

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Engineering Fluid Mechanics
© 2012 T. Al-Shemmeri & bookboon.com
ISBN 978-87-403-0114-4

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Engineering Fluid Mechanics Contents

Contents

Notation 7

1 Fluid Statics 14
1.1 Fluid Properties 14
1.2 Pascal’s Law 22
1.3 Fluid-Static Law 22
1.4 Pressure Measurement 26
1.5 Centre of pressure & the Metacentre 31
1.6 Resultant Force and Centre of Pressure on a Curved Surface in a Static Fluid 37
1.7 Buoyancy 40
1.8 Stability of floating bodies 43
1.9 Tutorial problems 49

2 Internal Fluid Flow 51
2.1 Definitions 52
2.2 Conservation of Mass 54
2.3 Conservation of Energy 56

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Engineering Fluid Mechanics Contents

2.4 Flow Measurement 59
2.5 Flow Regimes 63
2.6 Darcy Formula 64
2.7 The Friction factor and Moody diagram 65
2.8 Flow Obstruction Losses 69
2.9 Fluid Power 70
2.10 Fluid Momentum 73
2.11 Tutorial Problems 80

3 External Fluid Flow 82
3.1 Regimes of External Flow 82
3.2 Drag Coefficient 83
3.3 The Boundary Layer 85
3.4 Worked Examples 88
3.5 Tutorial Problems 97

4 Compressible Fluid Dynamics 99
4.1 Compressible flow definitions 99
4.2 Derivation of the Speed of sound in fluids 100
4.3 The Mach number 102

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Engineering Fluid Mechanics Contents

4.4 Compressibility Factor 105
4.5 Energy equation for frictionless adiabatic gas processes 109
4.6 Stagnation properties of compressible flow 113
4.7 Worked Examples 116
4.8 Tutorial Problems – Compressible Flow 121

5 Hydroelectric Power 123
5.1 Introduction 123
5.2 Types of hydraulic turbines 124
5.3 Performance evaluation of Hydraulic Turbines 127
5.4 Pumped storage hydroelectricity 130
5.5 Worked Examples 134
5.7 Tutorial Problems 137

Sample Examination paper 138

Formulae Sheet 146

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Engineering Fluid Mechanics Preface


Notation

Symbol definition units

A area m2

D diameter m

F force N

g gravitational acceleration m/s2

h head or height m

L length m

m mass kg

P pressure Pa or N/m2

∆P pressure difference Pa or N/m2

Q volume flow rate m3/s

r radius m

t time s

V velocity m/s

z height above arbitrary datum m

Subscripts

a atmospheric
c cross-sectional
f pipe friction
o obstruction
p pump
r relative
s surface
t turbine
x x-direction
y y-direction
z elevation

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Engineering Fluid Mechanics Preface


Dimensionless numbers

Cd discharge coefficient
f friction factor (pipes)
K obstruction loss factor
k friction coefficient (blades)
Re Reynolds number

Greek symbols

θ, α, φ angle degrees

µ dynamic viscosity kg/ms

ν kinematics viscosity m2/s

ρ density kg/m3

τ shear stress N/m2

η efficiency %

Dimensions and Units

Any physical situation, whether it involves a single object or a complete system, can be described in terms
of a number of recognisable properties which the object or system possesses. For example, a moving
object could be described in terms of its mass, length, area or volume, velocity and acceleration. Its
temperature or electrical properties might also be of interest, while other properties – such as density
and viscosity of the medium through which it moves – would also be of importance, since they would
affect its motion. These measurable properties used to describe the physical state of the body or system
are known as its variables, some of which are basic such as length and time, others are derived such
as velocity. Each variable has units to describe the magnitude of that quantity. Lengths in SI units are
described in units of meters. The “Meter” is the unit of the dimension of length (L); hence the area will
have dimension of L2, and volume L3. Time will have units of seconds (T), hence velocity is a derived
quantity with dimensions of (LT-1) and units of meter per second. A list of some variables is given in
Table 1 with their units and dimensions.

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Engineering Fluid Mechanics Preface


Definitions of Some Basic SI Units
Mass: The kilogram is the mass of a platinum-iridium cylinder kept at Sevres in France.

Length: The metre is now defined as being equal to 1 650 763.73 wavelengths in vacuum of the
orange line emitted by the Krypton-86 atom.

Time: T he second is defined as the fraction 1/31 556 925.975 of the tropical year for 1900.
The second is also declared to be the interval occupied by 9 192 631 770 cycles of the
radiation of the caesium atom corresponding to the transition between two closely
spaced ground state energy levels.

Temperature: T he Kelvin is the degree interval on the thermodynamic scale on which the temperature
of the triple point of water is 273.16 K exactly. (The temperature of the ice point is
273.15 K).

Definitions of Some Derived SI Units

Force:
The Newton is that force which, when acting on a mass of one kilogram gives it an acceleration of one
metre per second per second.

Work Energy, and Heat:

The joule is the work done by a force of one Newton when its point of application is moved through a
distance of one metre in the direction of the force. The same unit is used for the measurement of every
kind of energy including quantity of heat.

The Newton metre, the joule and the watt second are identical in value. It is recommended that the
Newton is kept for the measurement of torque or moment and the joule or watt second is used for
quantities of work or energy.

Quantity Unit Symbol
Length [L] Metre m
Mass [m] Kilogram kg
Time [ t ] Second s
Electric current [ I ] Ampere A
Temperature [ T ] degree Kelvin K
Luminous intensity [ Iv ] Candela cd

Table 1: Basic SI Units

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Quantity Unit Symbol Derivation
Force [ F ] Newton N kg-m/s2
Work, energy [ E ] J N-m
Power [ P ] joule W J/s
Pressure [ p ] watt Pa N/m2
Pascal

Table 2: Derived Units with Special Names

Quantity Symbol
Area m2
m3
Volume
Density kg/m3
Angular acceleration rad/s2
Velocity m/s
Pressure, stress N/m2
Kinematic viscosity m2/s
Dynamic viscosity N-s/m2
Momentum kg-m/s
Kinetic energy kg-m2/s2
Specific enthalpy J/kg
Specific entropy J/kg K

Table 3: Some Examples of Other Derived SI Units

Quantity Unit Symbol Derivation
Time minute min 60 s
Time h 3.6 ks
hour oC
Temperature degree Celsius K – 273.15
Angle o π/180 rad
Volume degree 10-3 m3 or dm3
Speed litre l
km/h -
Angular speed kilometre per hour rev/min -
Frequency revolution per minute cycle/s
Pressure Hz 102 kN/m2
hertz b 100 mm2/s
Kinematic viscosity bar St 100 mN-s/m2
Dynamic viscosity stoke P
poise
Table 4: Non-SI Units

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Name Symbol Factor Number
exa E 1018 1,000,000,000,000,000,000
Peta P 1015
tera T 1012 1,000,000,000,000,000
giga G 109 1,000,000,000,000
mega M 106 1,000,000,000
kilo k 103 1,000,000
hecto h 102 1,000
deca da 10 100
deci d 10-1 10
centi c 10-2 0.1
milli m 10-3 0.01
micro µ 10-6 0.001
nano n 10-9 0.000001
pico p 10-12 0.000000001
f 10-15 0.000000000001
fempto a 10-18 0.000000000000001
atto
0.000000000000000001
Table 5: Multiples of Units

item conversion
Length
Mass 1 in = 25.4 mm
Area 1 ft = 0.3048 m
1 yd = 0.9144 m
Volume 1 mile = 1.609 km
Force, Weight
Density 1 lb. = 0.4536 kg (0.453 592 37 exactly)
Velocity
1 in2 = 645.2 mm2
1 ft2 = 0.092 90 m2
1 yd2 = 0.8361 m2
1 acre = 4047 m2

1 in3 = 16.39 cm3
1 ft3 = 0.028 32 m3 = 28.32 litre
1 yd3 = 0.7646 m3 = 764.6 litre
1 UK gallon = 4.546 litre
1 US gallon = 3.785 litre

1 lbf = 4.448 N

1 lb/ft3 = 16.02 kg/m3

1 km/h = 0.2778 m/s
1 ft/s = 0.3048 m/s
1 mile/h = 0.4470 m/s = 1.609 km/h

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Pressure, Stress 1000 Pa = 1000 N/m2 = 0.01 bar
Power 1 in H2O = 2.491 mb
Moment, Torque 1 lbf/in2 (Psi)= 68.95 mb or 1 bar = 14.7 Psi
Rates of Flow
Fuel Consumption 1 horsepower = 745.7 W
Kinematic Viscosity
Dynamic Viscosity 1 ft-pdl = 42.14 mN-m

Energy 1 gal/h = 1.263 ml/s = 4.546 l/h
1 ft3/s = 28.32 l/s

1 mile/gal = 0.3540 km/l

1 ft2/s = 929.0 cm2/s = 929.0 St

1 lbf-s/ft2 = 47.88 N-s/m2 = 478.8 P
1 pdl-s/ft2 = 1.488 N-s/m2 = 14.88 P
1cP = 1 mN-s/m2

1 horsepower-h = 2.685 MJ
1 kW-h = 3.6 MJ
1 Btu = 1.055 kJ
1 therm = 105.5 MJ

Table 6: Conversion Factors

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Length (L) Unit X Factor = Unit x Factor = Unit

Area (A) ins 25.4 mm 0.0394 ins
ft 0.305 m 3.281 ft
Volume (V)
in2 645.16 mm2 0.0016 in2
Mass (M) ft2 0.093 m2 10.76 ft2
Force (F)
Velocity (V) in3 16.387 mm3 0.000061 in3
ft3 0.0283 m3 35.31 ft3
Volume Flow ft3 28.32 litre 0.0353 ft3
pints 0.5682 litre 1.7598 pints
Pressure (P) Imp. gal 4.546 litre 0.22 Imp gal
Density (ρ) Imp. gal 0.0045 m3 220 Imp gal
Heat Flow
Rate lb. 0.4536 kg 2.2046 lb.
Thermal tonne 1000 kg
Conductivity (k) 0.2248 lb.
Thermal lb. 4.448 N
Conductance (U) 196.85 ft/min
Enthalpy ft/min 0.0051 m/sec
(h) 13.2 Imp gal/min
Imp gal/min 0.0758 litres/s 7,936.5 Imp gal/h
Imp gal/h 0.00013 m3/s 2,118.6 ft3/min
ft3/min 0.00047 m3/s
14.5 lb/in2
lb/in2 0.0689 bar 1.02 kg/cm2
kg/cm2 0.9807 bar
0.0624 lb/ft3
lb/ft3 16.019 kg/m3
3.412 Btu/h
Btu/h 0.2931 W 0.8598 kcal/h
kcal/h 1.163 W
0.5777 Btu/ft h R
Btu/ft h R 1.731 W/m K 0.8598 kcal/m h K
kcal/m h K 1.163 W/m K
0.1761 Btu/h ft2 R
Btu/h ft2 R 5.678 W/m2 K 0.8598 kcal/h m2 K
kcal/h m2 K 1.163 W/m2 K
0.00043 Btu/lb.
Btu/lb. 2,326 J/kg 0.00024 kcal/kg
kcal/kg 4,187 J/kg

Table 7: Conversion Factors

Simply multiply the imperial by a constant factor to convert into Metric or the other way around.

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Engineering Fluid Mechanics Fluid Statics

1 Fluid Statics



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1.1 Fluid Properties

Fluid

A fluid is a substance, which deforms when subjected to a force. A fluid can offer no permanent resistance
to any force causing change of shape. Fluid flow under their own weight and take the shape of any solid
body with which they are in contact. Fluids may be divided into liquids and gases. Liquids occupy
definite volumes. Gases will expand to occupy any containing vessel.

S.I Units in Fluids

The dimensional unit convention adopted in this course is the System International or S.I system. In this
convention, there are 9 basic dimensions. The three applicable to this unit are: mass, length and time.
The corresponding units are kilogrammes (mass), metres (length), and seconds (time). All other fluid
units may be derived from these.

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Density

The density of a fluid is its mass per unit volume and the SI unit is kg/m3. Fluid density is temperature
dependent and to a lesser extent it is pressure dependent. For example the density of water at sea-level
and 4oC is 1000 kg/m3, whilst at 50oC it is 988 kg/m3.

The relative density (or specific gravity) is the ratio of a fluid density to the density of a standard reference
fluid maintained at the same temperature and pressure:

For gases: RDgas = ρgas = ρgas / m3
For liquids: ρair 1.205 kg

RDliquid = ρ liquid = ρ liquid m3
ρ water 1000 kg /

Viscosity

Viscosity is a measure of a fluid’s resistance to flow. The viscosity of a liquid is related to the ease with
which the molecules can move with respect to one another. Thus the viscosity of a liquid depends on the:

• Strength of attractive forces between molecules, which depend on their composition, size,
and shape.

• The kinetic energy of the molecules, which depend on the temperature.

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Viscosity is not a strong function of pressure; hence the effects of pressure on viscosity can be neglected.
However, viscosity depends greatly on temperature. For liquids, the viscosity decreases with temperature,
whereas for gases, the viscosity increases with temperature. For example, crude oil is often heated to a
higher temperature to reduce the viscosity for transport.

Consider the situation below, where the top plate is moved by a force F moving at a constant rate of
V (m/s).

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9HORFLW\ GY

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The shear stress τ is given by:

τ = F/A

The rate of deformation dv (or the magnitude of the velocity component) will increase with distance
above the fixed plate. Hence:

τ = constant x (dv / dy)

where the constant of proportionality is known as the Dynamic viscosity (µ) of the particular fluid
separating the two plates.

τ = µ x (V / y)

Where V is the velocity of the moving plate, and y is the distance separating the two plates. The units
of dynamic viscosity are kg/ms or Pa s. A non-SI unit in common usage is the poise where 1 poise =
10-1 kg/ms

Kinematic viscosity (ν) is defined as the ratio of dynamic viscosity to density.

i.e. ν = μ / ρ (1.1)

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The units of kinematic viscosity are m2/s.
Another non-SI unit commonly encountered is the “stoke” where 1 stoke = 10-4 m2/s.

Typical liquid Dynamic Viscosity Centipoise* (cp) Kinematic Viscosity Centistokes (cSt)
Water 1 1
Vegetable oil 34.6 43.2
SAE 10 oil 88 110
SAE 30 oil 352 440
Glycerine 880 1100
SAE 50 oil 1561 1735
SAE 70 oil 17,640 19,600

Table 1.1 Viscosity of selected fluids at standard temperature and pressure
Note: 1 cp = 10-3kg/ms and 1cSt = 10-6 m2/s

Figure 1.1 Variation of the Viscosity of some common fluids with temperature

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Worked Example 1.1
The temperature dependence of liquid viscosity is the phenomenon by which liquid viscosity tends to
decrease as its temperature increases. Viscosity of water can be predicted with accuracy to within 2.5%
from 0°C to 370°C by the following expression:

μ (kg/ms)= 2.414*10^-5 * 10^(247.8 K/(Temp – 140 K))

Calculate the dynamic viscosity and kinematic viscosity of water at 20oC respectively. You may assume
that water is incompressible, and its density is 1000 kg/m3.

Compare the result with that you find from the viscosity chart and comment on the difference.

Solution
a) Using the expression given:

μ (kg/ms) = 2.414*10 -5 * 10(247.8 K/(Temp – 140 K))

= 2.414x10-5x10(247.8/(20+273-140)

= 1.005x10-3 kg/ms

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Kinematic viscosity = dynamic viscosity / density

= 1.005x10-3/1000 = 1.005x10-6 m2/s

b) From the kinematic viscosity chart, for water at 20 is 1.0x10-6 m2/s.

The difference is small, and observation errors may be part of it.

Worked Example 1.2

A shaft 100 mm diameter (D) runs in a bearing 200 mm long (L). The two surfaces are separated by an
oil film 2.5 mm thick (c). Take the oil viscosity (µ) as 0.25 kg/ms. if the shaft rotates at a speed of (N)
revolutions per minute.

a) Show that the torque exerted on the bearing is given as:

7RUTXH P[S [ 1[/ ['
[F


b) Calculate the torque necessary to rotate the shaft at 600 rpm.

Solution:

a) The viscous shear stress is the ratio of viscous force divided by area of contact

W )
$

) P 9 F [$

$ S ' /

9 S'1

7RUTXH )[U P [S['[1 [ S ' / ['
[F

7RUTXH P [S [1[/ ['
[F

b) the torque at the given condition is calculated using the above equation:

7RUTXH P [S [1[/ [' [S [ [ [ 1P
[F [

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Engineering Fluid Mechanics Fluid Statics

Fluid Pressure

Fluid pressure is the force exerted by the fluid per unit area. Fluid pressure is transmitted with equal
intensity in all directions and acts normal to any plane. In the same horizontal plane the pressure
intensities in a liquid are equal. In the SI system the units of fluid pressure are Newtons/m2 or Pascals,
where 1 N/m2 = 1 Pa.

i.e. P = F (1.2)
A

Many other pressure units are commonly encountered and their conversions are detailed below:-

1 bar =105 N/m2

1 atmosphere = 101325 N/m2

1 psi (1bf/in2 – not SI unit) = 6895 N/m2

1 Torr = 133.3 N/m2

Terms commonly used in static pressure analysis include:

Pressure Head. The pressure intensity at the base of a column of homogenous fluid of a given height
in metres.

Vacuum. A perfect vacuum is a completely empty space in which, therefore the pressure is zero.

Atmospheric Pressure. The pressure at the surface of the earth due to the head of air above the surface.
At sea level the atmospheric pressure is about 101.325 kN/m2 (i.e. one atmosphere = 101.325 kN/m2 is
used as units of pressure).

Gauge Pressure. The pressure measured above or below atmospheric pressure.

Absolute Pressure. The pressure measured above absolute zero or vacuum.

Absolute Pressure = Gauge Pressure + Atmospheric Pressure (1.3)

Vapour Pressure

When evaporation of a liquid having a free surface takes place within an enclosed space, the partial
pressure created by the vapour molecules is called the vapour pressure. Vapour pressure increases
with temperature.

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Compressibility
A parameter describing the relationship between pressure and change in volume for a fluid.

A compressible fluid is one which changes its volume appreciably under the application of pressure.
Therefore, liquids are virtually incompressible whereas gases are easily compressed.

The compressibility of a fluid is expressed by the bulk modulus of elasticity (E), which is the ratio of the
change in unit pressure to the corresponding volume change per unit volume.

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Engineering Fluid Mechanics Fluid Statics

Solution:

H D E
I

)+

G

F

)9
Horizontal component = Force on horizontal projected area.

FH = ρ g hc A = 1000 x 9.81 x 0.75 x (3 x 1.5) = 33.1 x 103 N

Vertical component = weight of fluid which would occupy

Fv =ρg x vol of cylindrical sector = (1000 x 9.81) x (3 x π x 1.52) = 52 x 103 N
4

Resultant force )5 )+ )Y [ [ N1

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Engineering Fluid Mechanics Fluid Statics

If α is the angle of inclination of R to the horizontal then

tan α = FV/FH = 52 x 103/33.1 x 103 i.e. α = 57.28o

Since static pressure acts normal to the surface, it can be deduced that the line of action of FR passes
through the centre of curvature.

So the only force providing a moment is the weight of the gate.

Hence;

Moment = Force x distance = 6000 x 9.81 x 0.6 = 35300 Nm

1.7 Buoyancy

The buoyancy of a body immersed in a fluid is that property which will determine whether the body
will sink, rise or float. Archimedes established the analysis over 2000 years ago. Archimedes reasoned
that the volume of an irregular solid could be found by determining the apparent loss of weight when
the body is totally immersed in a liquid of known density.

Archimedes principle states:-

1. “The upthrust (vertical force) experienced by a body immersed in a fluid equals the weight of
the displaced fluid”

2. “A floating body displaces its own weight in the fluid in which it floats”.

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Upthrust

F = Pressure x Area

=PxA

But P = ρ.g.h

Therefore, F = ρ.g.h.A

But the volume VL= h.A
Therefore, F = ρ.g.VL (1.17)
Buoyant force can be expressed as:

F(b) = W(air) – W(liquid) = d x g x VL

where d is the density of the liquid, g is the acceleration of gravity and v is the volume of the immersed
object (or the immersed part of the body if it floats). Since W=mg, the apparent change in mass when
submerged is

m – m(apparent) = d(liquid) x vL

Worked Example 1.9
A hydrogen filled balloon has a total weight force of 9.5 kN. If the tension in the mooring cable anchoring
the balloon to the ground is 15.75 kN, determine the upthrust experienced by the balloon and its volume.

Take the density of air as 1.23 kg/m3.

Solution:

:

)
7

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Engineering Fluid Mechanics Fluid Statics

Since the system is stable: Upthrust = Weight force + Tension in cable

F = W + T

= 9.5 + 15.75

= 25.25 kN

The Upthrust is F = ρ x VL x g

Since the upthrust = the weight of displaced fluid, Therefore Balloon Volume

9/ ) [ P
U J [

Worked Example 1.10

A model boat consists of open topped rectangular metal can containing sand as a ballast. If the can has
a width of 100 mm, a length of 500 mm, and a mass of 1 kg, determine the mass of sand (kg) required
for the can to be immersed to a depth of 250 mm in sea water (RD = 1.03).

.

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Solution:

:HLJKW
:

)

Displaced volume VL = W x D x L = 0.1 x 0.25 x 0.5 = 0.0125 m3

For stable condition – Upthrust = weight force or F=W

The Upthrust due to Buoyancy = ρseawater g VL

The total weight = (mcan + msand) x 9.81

Therefore: ρseawater g VL. = (mcan + msand) x 9.81

1030 x 9.81 x 0.0125 = (1.0 + msand) x 9.81

Solving msand = 11.87 kg

Note: The sand will need to be levelled off or the can will not float vertically and may even be unstable.

1.8 Stability of floating bodies

A body is in a stable equilibrium if it returns to its original position after being slightly displaced. Neutral
position if the object remains in the new position after being slightly displaced. A body is in an unstable
equilibrium if it continues to move in the direction of the displacement.

D 6WDEOH FRQGLWLRQ E 1HXWUDO SRVLWLRQ F XQVWDEOH FRQGLWLRQ

Figure 1.6 Stability of floating objects

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Engineering Fluid Mechanics Fluid Statics

If the Centre of Buoyancy (B) is defined as the centre of gravity of the displaced fluid then the stability
of a floating object will depend on whether a righting or overturning moment is developed when the
centre of gravity (G) and the centre of buoyancy move out of vertical alignment due to the shifting of
the position of the latter. The centre of buoyancy moves because if a floating body tips, the shape of the
displaced liquid changes.

Position (a) Figure 1.7, illustrates a stable condition, where the forces of Buoyancy thrust and the weight
are equal and in line; while in Figure 1.7 (b) the body has been tipped over and the buoyancy has a
new position B, with G unchanged. The vertical through the new centre of buoyancy cuts the original
line, which is still passing through G at M, a point known as the Metacentre. In this case M lies above
G, and stability exists.

If M lies below G (c), it can be shown that once the body is tipped the couple introduced will aggravate the
rolling, causing it to tip further away from its stable position. The body is said to be unstable. Therefore,
for stability the metacentre must be above the centre of gravity, i.e. M above G.

Figure 1.7 Buoyancy and the metacentre

Worked Example 1.11
A raft floating in a river, supported by two drums, each 1m in diameter and 5m long.

If the raft is to stay afloat by 0.25m clear above water. What is the maximum weight that is allowed on it?
Assume density of water 1000 kg/m3.

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Solution:
The above case can be solved by first, calculating the displaced volume, converts it into a weight, and
then apply Archimedes’ principle

Fb = ρ g VL

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$ &
2

The angle AOC is calculated

cos(AOC) = 0.25/0.5

Hence angle AOC = 1.047 rad.

Area of sector = OC2 x angle = 0.52 x 1.047

= 0.262 m2

Area of triangle AOC $ [ P
Area submerged = 2 (0.262 – 0.108) = 0.308 m2

Volume displaced = 0.308 x 5 = 1.54 m3

Weight = Density x Volume displaced = 1000 x 1.54 = 1540 kg

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Worked Example 1.12

King Hero ordered a new crown to be made from pure gold (density = 19200 kg/m3). When he received
the crown he suspected that other metals may have been used in the construction. Archimedes discovered
that the crown needed a force of 20.91 N to suspend when submersed in water and that it displaced
3.1x10-4 m3 of water. He concluded that the crown could not be pure gold. Do you agree or disagree?

Solution:

¦ ) P D

)W )J )%

UV J 9 UZ J 9

KHQFH

UV UZ NJ P
[ [

The density of pure gold (19200 kg/m3) is more than twice this, so some other metal have been used,
such metal as steel. So agreed with Archimedes

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Worked Example 1.13

The hydraulic jack shown, the piston weighs 1000 N, determine the weight of the car which is supported
by the jack when the gauge reading is 1.2 bar. Assume that the jack cylinder has a diameter of 0.4 m.

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Engineering Fluid Mechanics Fluid Statics

Solution:

¦) R

3D $ )FDU )SLVWRQ

[ [ S [ )FDU
KHQFH


)FDU 1 NJ

ZHLJKW WRQQH

1.9 Tutorial problems

1.1 Show that the kinematic viscosity has the primary dimensions of L2T-1.

1.2 In a fluid the velocity measured at a distance of 75mm from the boundary is 1.125m/s. The fluid
has absolute viscosity 0.048 Pa s and relative density 0.913. What is the velocity gradient and shear
stress at the boundary assuming a linear velocity distribution? Determine its kinematic viscosity.

[Ans: 15 s-1, 0.72Pa.s; 5.257x10-5 m2/s]

1.3 A dead-weight tester is used to calibrate a pressure transducer by the use of known weights
placed on a piston hence pressurizing the hydraulic oil contained. If the diameter of the piston
is 10 mm, determine the required weight to create a pressure of 2 bars.

[Ans: 1.6 kg]

1.4 How deep can a diver descend in ocean water without damaging his watch, which will withstand
an absolute pressure of 5.5 bar?

Take the density of ocean water, = 1025 kg/m3.

[Ans: 44.75 m]

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EngEinngeienreinegrinFlguFidluMideMcheacnhiacsnics FluidFluSitdatSictas tics

1.51.5 TheTUhe-tuUb-etumbaenmomaentoermsehtoewr nshboewlonw,bperloovwe,thpartovtheetdhiafftetrhenecdeiifnfeprerenscseurienispgreivsesnurbeyi:s given by:

  d  2 
ρ.g.z2 1 +  D  
P1 − P2 = 


1.61.6 A flAat fcliartcucliarrcpullaatre,p1l.a2t5em, 1d.2ia5mmetedriiasmimetmererissediminmsewrsaegde winatseerw(daegnesiwtya1te2r00(dkegn/msi3t)ysu1c2h00thkatgi/tms g3r)esautecsht and

leastthdaetpitthssgarereat1e.s5t0amndanledas0t.6d0empthressapreect1iv.5e0ly.mDeatnedrm0i.n6e0tmherfeosrpceecetxievretleyd. Donetoenrme fiancee tbhyethfoerwceateexreprrteesdsure,

on one face by the water pressure, [Ans: (15180 N]

[Ans: (15180 N]
1.7 A rectangular block of wood, floats with one face horizontal in a fluid (RD = 0.9). The wood’s density is 750 kg/

1.7 m 3.ADeretecrtmaningeutlahre bpleorcceknotafgwe oofotdh,eflwooaotsd,wwithhichonisenfoact esuhbomreizrogendt.al in a fluid (RD = 0.9). The wood’s
density is 750 kg/m3. Determine the percentage of the wood, which is not submerged. [Ans: 17%]

1.8 An empty balloon and its equipment weight 50 kg, is inflated to a diameter of 6m, with a gas o[fAdennss:i1ty7%0.6] kg/
m3. What is the maximum weight of cargo that can be lifted on this balloon, if air density is assumed constant

1.8 a t 1A.2nkge/mmp3?ty balloon and its equipment weight 50 kg, is inflated to a diameter of 6m, with a gas
of density 0.6 kg/m3. What is the maximum weight of cargo that can be lifted on th[isAbnsa:ll1o7o.8n6, kg]
if air density is assumed constant at 1.2 kg/m3?
[Ans: 17.86 kg]

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